SSC CGL Previous Year Question Paper 24th August Shift 1 | QA | Quantitative Ability questions | CGL Past Year Papers

1. SSC CGL 24th August Shift 1 - QA | Simple Equation - SSC

If a⁴ + b⁴ + a²b² = 273 and a² + b² - ab = 21, then one of the values of $(\frac{1}{a} + \frac{1}{b})$ is:

A.
- $\frac{9}{4}$
B.
- $\frac{3}{4}$
C.
$\frac{9}{8}$
D.
$\frac{3}{2}$

2. SSC CGL 24th August Shift 1 - QA | Time, Speed & Distance - SSC

A train covers 450 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less to cover the same distance. How much time will it take to cover 315 km at its usual speed?

A.
7h 52m
B.
6h 30m
C.
6h 18m
D.
7h

Answer: Option D

Explanation :

Let the uniform speed of train = s

According to the problem,

$\frac{450}{s + 5}$ = $\frac{450}{s}$ - 1

$\frac{450}{s}$ - $\frac{450}{s + 5}$ = 1

450 $(\frac{s + 5 - s}{s (s + 5)})$ = 1

450 $(\frac{5}{s^{2} + 5 s})$ = 1

s² + 5s - 2250 = 0

(s + 50)(s − 45) = 0
s = -50 or s = 45

s cannot be negative, so s = 45 km/h

The uniform speed of train = 45 km/h

Time taken by train to cover 315 km at its usual speed = $\frac{315}{45}$ = 7 hours

Hence, the correct answer is Option D

3. SSC CGL 24th August Shift 1 - QA | Percentage, Profit & Loss - SSC

If selling price of 75 articles is equal to cost price of 60 articles, then the approximate loss or gain percent is:

A.
Profit of 25%
B.
No profit no loss
C.
Loss of 30%
D.
Loss of 20%

Answer: Option D

Explanation :

Let the cost price of 60 articles = C

Cost price of 1 article = $\frac{C}{60}$
Selling price of 75 articles = C

Selling price of 1 article = $\frac{C}{75}$

Loss% = $\frac{\frac{C}{60} - \frac{C}{75}}{\frac{C}{60}}$ × 100

= $\frac{\frac{5 C - 4 C}{300}}{\frac{C}{60}}$ × 100

= $\frac{60}{300}$ × 100

= 20%

Hence, the correct answer is Option D

4. SSC CGL 24th August Shift 1 - QA | Ratio, Proportion & Variation - SSC

If p is the third proportional to 3, 9, then what is the fourth proportional to 6, p, 4?

A.
$\frac{3}{2}$
B.
10
C.
18
D.
2 $\sqrt{3}$

Answer: Option C

Explanation :

p is the third proportional to 3, 9.

$\frac{p}{9}$ = $\frac{9}{3}$
p = 27

Let the 'n' is the fourth proportional of 6, p, 4.

n x 6 = p x 4

n x 6 = 27 x 4

n = 18

Fourth proportional of 6, p, 4 is 18.

Hence, the correct answer is Option C

5. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

Let △ABC ~ △RPQ and $\frac{a r (△ A B C)}{a r (△ P Q R)}$ = $\frac{16}{25}$ . If PQ = 4 cm, QR = 6 cm and PR = 7 cm, then AC (in cm) is equal to:

A.
6
B.
4.8
C.
3.6
D.
7.2

Answer: Option B

Explanation :

△ABC ~ △RPQ

⇒ ${(\frac{A C}{Q R})}^{2}$ = $\frac{a r (△ A B C)}{a r (△ P Q R)}$

⇒ ${(\frac{A C}{6})}^{2}$ = $\frac{16}{25}$

⇒ $\frac{A C}{6}$ = $\frac{4}{5}$

⇒ AC = 4.8 cm

Hence, the correct answer is Option B

6. SSC CGL 24th August Shift 1 - QA | Tables & Graphs - SSC

Study the following table and answer the question:
Number of students Appeared (A) and Passed (P) in an annual examination from four schools Q, R, S & T in five years (2014 to 2018)

The total number of students passed from school S in 2014 and 2017 is what percent of 90% of the total number of students appeared from school T in 2015, 2016 and 2017? (correct to one decimal place)

A.
53.9
B.
54.8
C.
59.6
D.
57.4

Answer: Option C

Explanation :

The total number of students passed from school S in 2014 and 2017 = 273 + 247 = 520

90% of the total number of students appeared from school T in 2015, 2016 and 2017 = $\frac{90}{100}$ (300 + 320 + 350)
= 873

Required percentage = $\frac{520}{873}$ × 100

= 59.6(approximately)

Hence, the correct answer is Option C

7. SSC CGL 24th August Shift 1 - QA | Number Theory - SSC

The sum of 3-digit numbers abc, cab and bca is not divisible by:

A.
37
B.
3
C.
31
D.
a + b + c

8. SSC CGL 24th August Shift 1 - QA | Percentage, Profit & Loss - SSC

A customer wanted to purchase an item marked for ₹10000. Shopkeeper offered two types of discounts, 25% flat discount or successive discounts of 14% and 12%. Which is the better offer for the customers and by how much?

A.
first offer by ₹68
B.
second offer by ₹68
C.
first offer by ₹32
D.
second offer by ₹100

Answer: Option A

Explanation :

25% flat discount

Selling price of the item = $\frac{75}{100}$ × 10000 = ₹7500
Successive discounts of 14% and 12%

Price of the item after 14% discount = $\frac{86}{100}$ × 10000 = ₹8600
Price of the item after 12% discount = $\frac{88}{100}$ × 8600 = ₹7568
Difference between selling prices = 7568 - 7500 = ₹68

First offer is better by ₹68.

Hence, the correct answer is Option A

9. SSC CGL 24th August Shift 1 - QA | Trigonometry - SSC

If tan θ + 3 cot θ - 2 $\sqrt{3}$ = 0, 0° < θ < 90°, then what is the value of (cosec² θ + cos² θ)?

A.
$\frac{2}{3}$
B.
$\frac{19}{12}$
C.
$\frac{14}{3}$
D.
$\frac{11}{12}$

Answer: Option B

Explanation :

tan θ + 3 cot θ - 2 $\sqrt{3}$ = 0

tan θ + $\frac{3}{\tan θ}$ - 2 $\sqrt{3}$ = 0

tan² θ - 2 $\sqrt{3}$ tan θ + 3 = 0

(tan θ - $\sqrt{3}$ )² = 0

tan θ - $\sqrt{3}$ = 0

tan θ = $\sqrt{3}$

0° < θ < 90°

⇒ θ = 60°

cosec² θ + cos² θ = cosec² 60° + cos² 60°

= ${(\frac{2}{\sqrt{3}})}^{2}$ + ${(\frac{1}{2})}^{2}$

= $\frac{4}{3}$ + $\frac{1}{4}$

= $\frac{16 + 3}{12}$

= $\frac{19}{12}$

Hence, the correct answer is Option B

10. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

ABCD is a cyclic quadrilateral such that when sides AB and DC are produced, they meet at E, and sides AD and BC meet at F, when produced. If ∠ADE = 80° and ∠AED = 50°, then what is the measure of ∠AFB?

A.
40°
B.
20°
C.
50°
D.
30°

11. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

In △ABC, right angled at B, if cot A = $\frac{1}{2}$ , then the value of $\frac{\sin A (\cos C + \cos A)}{\cos C (\sin C - \sin A)}$ is

A.
3
B.
-3
C.
-2
D.
2

Answer: Option B

Explanation :

cot A = $\frac{Adjacent side}{Opposite side}$ $\frac{1}{2}$

$\frac{\sin A (\cos C + \cos A)}{\cos C (\sin C - \sin A)}$ = $\frac{\frac{2}{\sqrt{5}} (\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}})}{\frac{2}{\sqrt{5}} (\frac{1}{\sqrt{5}} - \frac{2}{\sqrt{5}})}$

= $\frac{(\frac{3}{\sqrt{5}})}{(- \frac{1}{\sqrt{5}})}$

= -3

Hence, the correct answer is Option B

12. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

The area of a table top in the shape of an equilateral triangle is 9 $\sqrt{3}$ cm². What is the length (in cm) of each side of the table?

A.
6
B.
2
C.
4
D.
3

Answer: Option A

Explanation :

Let the length of each side of the table = a

The area of a table top in the shape of an equilateral triangle is 9 $\sqrt{3}$ cm².

$\frac{\sqrt{3}}{4}$ a² = 9 $\sqrt{3}$

a² = 36

a = 6 cm

Length of each side of the table = 6 cm

Hence, the correct answer is Option A

13. SSC CGL 24th August Shift 1 - QA | Tables & Graphs - SSC

Study the following table and answer the question:
Number of students Appeared (A) and Passed (P) in an annual examination from four schools Q, R, S & T in five years (2014 to 2018)

The difference between the average number of students passed from school R in 2015 to 2017 and the number of students passed from school Q in 2015 is x. The value of x lies between:

A.
20 and 25
B.
35 and 40
C.
30 and 35
D.
25 and 30

Answer: Option C

Explanation :

The difference between the average number of students passed from school R in 2015 to 2017 and the number of students passed from school Q in 2015 is x.

x = 320 - $(\frac{285 + 294}{2})$

x = 320 - 289.5

x = 30.5

x lies between 30 and 35.

Hence, the correct answer is Option C

14. SSC CGL 24th August Shift 1 - QA | Tables & Graphs - SSC

Table shows the number of trees planted in 4 cities from 2016 to 2020.

What is the total number of trees planted in Chandigarh in 2017 and in Kolkata in 2020?

A.
4500
B.
4750
C.
4250
D.
3550

15. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

The vertices of a △ABC lie on a circle with centre O. AO is produced to meetthe circle at the point P. D is a point on BC such that AD ⊥ BC. If ∠B = 68° and ∠C = 52°, then the measure of ∠DAP is:

A.
18°
B.
16°
C.
12°
D.
28°

16. SSC CGL 24th August Shift 1 - QA | Time & Work - SSC

A can complete a work in 11 $\frac{1}{2}$ days. B is 25% more efficient than A and C is 50% less efficient than B. Working together A, B and C will complete the same work

A.
5 days
B.
4 days
C.
3 days
D.
8 days

Answer: Option B

Explanation :

Let the total work = 460 units

A can complete a work in 11 $\frac{1}{2}$ days.

Efficiency of A = $\frac{460}{\frac{23}{2}}$ = 40 units/day

B is 25% more efficient than A.

Efficiency of B = $\frac{125}{100}$ × 40 = 50 units/day

C is 50% less efficient than B.

Efficiency of C = $\frac{50}{100}$ × 50 = 25 units/day

Efficiency of A, B and C together = 40 + 50 + 25 = 115 units/day

Number of days required for A, B and C together to complete the work = $\frac{460}{115}$ = 4 days

Hence, the correct answer is Option B

17. SSC CGL 24th August Shift 1 - QA | Average, Mixture & Alligation - SSC

The average of 23 numbers is 51. The average of first 12 numbers is 49 and the average of last 12 numbers is 54. If the twelfth number is removed, then the average of the remaining numbers (correct to two decimal places) is:

A.
53.25
B.
50.45
C.
51.75
D.
52.65

Answer: Option B

Explanation :

The average of 23 numbers is 51.

Sum of the 23 numbers = 51 x 23 = 1173.......(1)

The average of first 12 numbers is 49.

Sum of the first 12 numbers = 49 x 12 = 588

Sum of the first 11 numbers + 12th number = 588.....(2)

The average of last 12 numbers is 54.

Sum of the last 12 numbers = 54 x 12 = 648

12th number + Sum of the last 11 numbers = 648.....(3)

Adding (2) and (3),

Sum of the first 11 numbers + 12th number + 12th number + Sum of the last 11 numbers = 588 + 648

(Sum of the first 11 numbers + 12th number + Sum of the last 11 numbers) + 12th number = 1236

Sum of the 23 numbers + 12th number = 1236

1173 + 12th number = 1236

12th number = 63

Required average = $\frac{1173 - 63}{22}$

= 50.4545

Hence, the correct answer is Option B

18. SSC CGL 24th August Shift 1 - QA | Percentage, Profit & Loss - SSC

The total number of students in a school is 1400, out of which 35% of the students are girls and the rest are boys. If 80% of the boys and 90% of the girls passed in an annual examination, then the percentage of the students who failed is:

A.
16.5
B.
21.5
C.
17.4
D.
15.8

Answer: Option A

Explanation :

Total number of students in a school is 1400.

35% of the students are girls and the rest are boys.

Number of girls = $\frac{35}{100}$ × 1400 = 490

Number of boys = 1400 - 490 = 910

80% of the boys and 90% of the girls passed in an annual examination.

Number of students passed in the examination = $\frac{80}{100}$ × 910 + $\frac{90}{100}$ × 490 = 728 + 441 = 1169

Number of students failed in the examination = 1400 - 1169 = 231

Percentage of the students failed in the examination = $\frac{231}{1400}$ × 100

= 16.5%

Hence, the correct answer is Option A

19. SSC CGL 24th August Shift 1 - QA | Simple Equation - SSC

If x + $\frac{1}{x}$ = 7, then x² + $\frac{1}{x^{2}}$ is equal to:

A.
47
B.
49
C.
61
D.
51

Answer: Option A

Explanation :

x + $\frac{1}{x}$ = 7

Squaring on both sides,

x² + $\frac{1}{x^{2}}$ + 2.x. $\frac{1}{x}$ = 49

x² + $\frac{1}{x^{2}}$ + 2 = 49

x² + $\frac{1}{x^{2}}$ = 47

Hence, the correct answer is Option A

20. SSC CGL 24th August Shift 1 - QA | Triangles - SSC

In a triangle ABC, length of the side AC is 4 cm more than 2 times the length of the side AB. Length of the side BC is 4 cm less than the three times the length of the side AB. If the perimeter of △ABC is 60 cm, then its area (in cm²) is:

A.
150
B.
144
C.
100
D.
120

Answer: Option D

Explanation :

Length of the side AC is 4 cm more than 2 times the length of the side AB.

b = 2c + 4......(1)

Length of the side BC is 4 cm less than the three times the length of the side AB.

a = 3c - 4.......(2)

Perimeter of △ABC is 60 cm.

a + b + c = 60

3c - 4 + 2c + 4 + c = 60

6c = 60

c = 10 cm

b = 2c + 4 = 24 cm

a = 3c - 4 = 26 cm

Half of the perimeter(s) = 30 cm

Area of the triangle = $\sqrt{s (s - a) (s - b) (s - c)}$

= $\sqrt{30 (30 - 26) (30 - 24) (30 - 10)}$

= $\sqrt{120 \times 120}$

= 120 cm²

Hence, the correct answer is Option D

21. SSC CGL 24th August Shift 1 - QA | Simple Equation - SSC

If 2x + 3y + 1 = 0, then what is the value of (8x³ + 8 + 27y³ - 18xy)?

A.
-7
B.
7
C.
-9
D.
9

22. SSC CGL 24th August Shift 1 - QA | Percentage, Profit & Loss - SSC

A sum of ₹9500 amounts to ₹11495 in 2 years at a certain rate percent per annum, interest compounded yearly. What is the simple interest (in ₹) on the same sum for the same time and double the rate?

A.
3990
B.
3420
C.
4560
D.
3800

Answer: Option D

Explanation :

Let the rate of interest = R

11495 = 9500 ${(1 + \frac{R}{100})}^{2}$

${(1 + \frac{R}{100})}^{2}$ = $\frac{11495}{9500}$

${(1 + \frac{R}{100})}^{2}$ = $\frac{121}{100}$

1 + $\frac{R}{100}$ = $\frac{11}{10}$

$\frac{R}{100}$ = $\frac{1}{10}$

R = 10%

Simple interest = $\frac{9500 \times 2 \times 20}{100}$

= ₹3800

Hence, the correct answer is Option D

23. SSC CGL 24th August Shift 1 - QA | Tables & Graphs - SSC

Table shows income (in ₹) received by 4 employees of a company during the month of December 2020 and all their income sources.

What is the ratio of salary of Varun to his income other than salary?

A.
128 : 653
B.
653 : 128
C.
128 : 525
D.
525 : 128

24. SSC CGL 24th August Shift 1 - QA | Trigonometry - SSC

Simplify:

sec² α $(1 + \frac{1}{\cos e c α})$ $(1 - \frac{1}{\cos e c α})$ .

A.
tan⁴ α
B.
sin² α
C.
1
D.
-1

Answer: Option C

Explanation :

sec² α $(1 + \frac{1}{\cos e c α})$ $(1 - \frac{1}{\cos e c α})$ = $\frac{1}{\cos^{2} α}$ (1 + sin α) (1 - sin α)

= $\frac{1}{\cos^{2} α}$ (1 - sin² α)

= $\frac{1}{\cos^{2} α}$ (cos² α)

= 1

Hence, the correct answer is Option C

25. SSC CGL 24th August Shift 1 - QA | Simplification - SSC

The value of

3 $\frac{1}{5}$ ÷ 4 $\frac{1}{2}$ of 5 $\frac{1}{3}$ - $\frac{1}{8}$ ÷ $\frac{1}{2}$ of $\frac{1}{4}$ + $\frac{1}{4}$ $(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4})$ is:

A.
- $\frac{37}{60}$
B.
- $\frac{17}{60}$
C.
$\frac{17}{60}$
D.
$\frac{37}{60}$

Answer: Option A

Explanation :

3 $\frac{1}{5}$ ÷ 4 $\frac{1}{2}$ of 5 $\frac{1}{3}$ - $\frac{1}{8}$ ÷ $\frac{1}{2}$ of $\frac{1}{4}$ + $\frac{1}{4}$ $(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4})$

= $\frac{16}{5}$ ÷ $\frac{9}{2}$ of $\frac{16}{3}$ - $\frac{1}{8}$ ÷ $\frac{1}{2}$ of $\frac{1}{4}$ + $\frac{1}{4}$ $(\frac{1}{2} \times \frac{8}{1} \times \frac{1}{4})$

= $\frac{16}{5}$ ÷ 24 - $\frac{1}{8}$ ÷ $\frac{1}{8}$ + $\frac{1}{4}$ (1)

= $\frac{16}{5 \times 24}$ - 1 + $\frac{1}{4}$

= $\frac{8 - 60 + 15}{60}$

= - $\frac{37}{60}$

Hence, the correct answer is Option A

SSC CGL 24th August Shift 1 - QA

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